Question

Professor Van Eggmon thinks that dissolving eggshells at 74.1o F is easier than cracking eggs. The shells are made of calcium carbonate (CaCO3), which reacts with acetic acid (CH3COOH) to form carbon dioxide (CO2), water (H2O), and calcium acetate (Ca(CH3COO)2). The feed to the egg dissolver (reactor) contains 46.2 mol/s calcium carbonate and 352 mol/s dilute acetic acid solution (4.15 mol% acetic acid and the balance water). The conversion of acetic acid is 69.4%. a. (5 points) Draw and label a process flow diagram. Number the streams and clearly label the components in each stream. b. (3 points) Write the limiting reactant in the box. c. (34 points) Determine the component flow rates (mol/s) exiting the reactor. d. (3 points) Find the temperature in Kelvin (K). e. (3 points) The professor’s children state that the feed flow rate of CaCO3 of 9.43 x 107 g/day. Compared to the original operating conditions, the feed flow rate of CaCO3 will (circle one): INCREASE DECREASE STAY THE SAME

Answer 1

(b) Given feed flow rate of CaCO3 = 46.2 mol/s

flow rate of CH3COOH = 352 mol/s x (4.15/100) = 14.6 mol/s

flow rate of water = 352 - 14.6 = 337.4 mol/s

Given the conversion rate of CH3COOH = 69.4 % = 14.6 x (69.4/100) = 10.1 mol/s

Hence only 10.1 mol/s of CH3COOH reacts during the reaction.

Consider 1 second of the reaction

The chemical reaction is

-----------------CaCO3(s) + 2CH3COOH(aq) ---- > Ca(CH3COO)2 + CO2 + H2O

Initial moles:46.2 mol ----10.1 mol ------------------- 0 mol ------------- 0 mol, 337.4 mol

moles aft.rxn:(46.2-5.05), (10.1-10.1) ----------------5.05 mol, --------5.05 mol,(337.4+5.05) mol

----------------- = 41.2 mol, 0 mol --------------------- 5.05 mol ---------5.05 mol, 342.5 mol

In the above reaction CH3COOH is completely used up. Hence CH3COOH is the limiting reactant.

(c) The component flow rates (mol/s) exiting the reactor are

CaCO3 = 41.2 mol/s (answer)

CH3COOH = 14.6x(100 - 69.4)/100 = 4.5 mol/s (answer)

Ca(CH3COO)2 = 5.05 mol/s (answer)

CO2 = 5.05 mol/s (answer)

H2O = 342.5 mol/s (answer)

(d): Temperature in kelvin = 296.5 K

Original feed flow rate of CaCO3 = 46.2 mol/s = 46.2 x 60 x 60 x 24 mol x (100.0 g/mol) = 3.99x10⁸ g/day

professor’s children's feed flow rate = 9.43x10⁷ g/day

Hence compared to original operating condition feed flow rate decreases (answer)

(a) Process flow diagram: